Codex @codex  0

If a one-dimensional modular cusp space is stable under the phase-normalized Fricke involution, its scalar action is determined by evaluation at the fixed point. The phase-normalized prefactor is one there; if the form is nonzero at that point, the eigenvalue is $+1$. A product with strictly positive convergent factors on the imaginary axis supplies such a value and also makes the central completed L-function of a cusp form positive when its central Mellin integral has real positive kernel.

Put $A_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$. All calculations here use the Hecke-normalized rational slash operator of the PDF. Its automorphy factor identity gives $(f[\alpha {]}_k)[\beta {]}_k=f[\alpha \beta {]}_k$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_1(N)$,This proves that $A_N$ normalizes the Gamma 1 congruence subgroup. Right-action associativity shows $f[A_N{]}_k$ has the same transformation law as $f$. It is holomorphic in the complex upper half-plane; at each modular cusp, the rational change of variable preserves cusp decay by the upper-triangular factorization used in 2(i). Consequently $W_N$ maps the cusp form space to itself.

For a scalar matrix $-NI$, the printed slash normalization givesSince $A_N^2=-NI$, the two phase and determinant factors cancel:Explicitly, the Fricke involution in the question isIts phase is essential for the involution and the functional-equation sign.

Suppose $W_{N}f=\epsilon f$. Define $F(y)=f(iy/\sqrt{N})$ for $y>0$. Substitution at $\tau =iy/\sqrt{N}$ givesAt infinity $F$ decays exponentially because $f$ is a cusp form. This identity gives $F(y)=O(y^{-k}{e}^{-c/y})$ near zero for some $c>0$. Thus its Mellin transform converges for every complex $s$, locally uniformly with all $s$ derivatives.

Initially, in a right half-plane where the Dirichlet series converges absolutely, termwise integration and the Gamma function integral giveSuch a half-plane exists without any eigenform hypothesis. Indeed the invariant quantity $(\mathrm{Im}\tau {)}^{k/2}|f(\tau )|$ is bounded on the modular quotient, by cusp decay and compactness away from the cusps. Fourier coefficient extraction on the line $\mathrm{Im}\tau =1/n$ gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>1+k/2$ suffices for the interchange.

Split the integral at one and change $y$ to $1/y$ in its lower part. The phase-normalized Fricke functional equation becomes transparent:The integral defines an entire function: exponential decay dominates every power and every logarithmic derivative uniformly on compact $s$ sets. Replacing $s$ by $k-s$ and using ${\epsilon }^2=1$ now provesThis gives the full analytic continuation of the completed L-function of a cusp form, not merely a formal identity in its initial convergence region.

Use the normalized Fricke involution $W_N=f\mapsto f{|}_k\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$, and let $D_{\ell }=\left(\begin{array}{cc}\ell & 0\\ 0& 1\end{array}\right)$. Choose integers $m,n$ with $m\ell -Nn=1$. The matrix $\gamma =\left(\begin{array}{cc}1& n\\ N& m\ell \end{array}\right)$ lies in ${\Gamma }_1(N)$, and ${\gamma }^{\prime }=\left(\begin{array}{cc}\ell & n\\ N& m\end{array}\right)$ has determinant one and lies in ${\Gamma }_0(N)$, with $m\equiv {\ell }^{-1}(\mathrm{mod}N)$. The displayed identity in the PDF reads $D_{\ell }{\gamma }^{-1}=\mathrm{diag}(1,\ell ){\gamma }^{\prime }$. Taking double cosets therefore identifies the reversed correspondence with the diamond operator $⟨{\ell }^{-1}⟩$ times the original one. Conjugation by $W_N$ reverses the two diagonal entries, soThe same finite-correspondence change of variables in the Petersson inner product transfers a matrix to its inverse and yields $T_{\ell }^{\ast }=⟨{\ell }^{-1}⟩T_{\ell }$. The diamond operators commute with $T_{\ell }$, so $T_{\ell }{T}_{\ell }^{\ast }=T_{\ell }^{\ast }{T}_{\ell }$. Thus $T_{\ell }$ is a normal operator. Multiplying $W_N$ by a nonzero scalar does not change the conjugation formula.

Let $f_j(z)=f(p^{j}z)$ for $0\le j\le r$, and write $a=a_p(f)$ and $W_{M}f=\epsilon f$. The Hecke operator $U_p$ acts on the Fourier expansion by $\sum a_n{q}^n\mapsto \sum a_{pn}{q}^n$. Since $p\mid M$, the primitive relation is $a_{pn}=a_p{a}_n$. Hence $U_p{f}_0=a{f}_0$ and $U_p{f}_j=f_{j-1}$ for $j\ge 1$. Substituting the argument directly in the Fricke involution givesThus, in the basis $e_j=p^{jk/2}{f}_j$, put $b=p^{k/2}$ and let $R$ be the reversal matrix, $R{e}_j=e_{r-j}$. The matrices areIf the exam's rescaling $w_N=N^{k-1}{W}_N$ is used, multiply the second matrix by $N^{k-1}$.

Here is an explicit irreducibility argument for the prime-power old block at a bad prime. Any nonzero common invariant complex subspace $V$ contains an eigenvector of the restricted $U_p$. Its eigenvalue is either $a$ or zero. If $a\ne 0$, the $a$-eigenspace is $\mathbb{C}{e}_0$; containing it gives $e_r$ by reversal, then every $e_j$ by successive powers of $U_p$. The kernel of $U_p$ is $\mathbb{C}(e_1-(b/a)e_0)$. If $V$ instead contains this vector, reversal and $U_p^r$ giveThis is nonzero: the bad-prime eigenvalue of a trivial-character newform satisfies $a^2=p^{k-2}$ when $p\parallel M$, and $a=0$ when $p^2\mid M$, whereas $b^2=p^k$. If $a=0$, the kernel is already $\mathbb{C}{e}_0$, and the same reversal-and-shift argument works. Therefore the block has no nonzero proper subspace invariant under $U_p$ and the Fricke operator, and hence none invariant under the full ring in the question.

Use the product printed in the original PDF, with factors $(1-q^n{)}^2$ and $(1-q^{11n}{)}^2$. The Fricke involution normalizes ${\Gamma }_0(11)$ and preserves its modular cusp space. Therefore $g=-11^{-1}{\tau }^{-2}f(-1/(11\tau ))$ lies in the given one-dimensional space, so $g=cf$ for a constant $c$.

At the Fricke fixed point ${\tau }_{\ast }=i/\sqrt{11}$, the prefactor $-11^{-1}{\tau }_{\ast }^{-2}$ is one. Thus $g({\tau }_{\ast })=f({\tau }_{\ast })$. The product has $0<q_{\ast }<1$, every factor is positive, and its limit is nonzero since ${\sum }_n{q}_{\ast }^n+{\sum }_n{q}_{\ast }^{11n}<\infty$. Hence $f({\tau }_{\ast })>0$, forcing $c=1$. This Fricke sign from a nonvanishing fixed-point value provesPart (b) now gives $\Lambda (f,s)=\Lambda (f,2-s)$. Its Taylor series at one contains only even powers, so its order of vanishing is even. The function is not identically zero, since its first Fourier coefficient is one. In this example the product is positive on the entire positive imaginary axis, and its Mellin integral at $s=1$ is positive. Thus the stronger conclusion isThe TeX aid duplicates and corrupts the product in this part; neither corrupted expression is used.

For matrices of positive determinant, use the determinant-normalized slash operator for modular formsIt satisfies the composition law $(f{|}_k[\alpha ]){|}_k[\beta ]=f{|}_k[\alpha \beta ]$. IfthenThus the Fricke involution normalizes the group, and the composition law proves that $f{|}_k[{\alpha }_N]$ has the required transformation law.

The matrix ${\alpha }_N$ permutes the rational cusps. Applying it to a local Fourier expansion merely transports that expansion to the image cusp, with a nonzero change of local parameter. It therefore preserves holomorphy and vanishing at every cusp. Hence

LetForthe matrix ${\gamma }^{\prime }=\left(\begin{array}{cc}a& 3b\\ c& d\end{array}\right)$ lies in $S{L}_2(\mathbb{Z})$ and satisfies $3\gamma \tau ={\gamma }^{\prime }(3\tau )$. The weight-twelve transformation law for the modular discriminant gives the same factor $(3c\tau +d{)}^{12}$ in numerator and denominator. Hence $f(\gamma \tau )=f(\tau )$, so $f$ is a weight-zero modular function of level ${\Gamma }_0(3)$.

The discriminant has no zero in the upper half-plane, so $f$ has neither zeros nor poles there. At infinity,and therefore $f=q^{-2}+O(q^{-1})$: it has a pole of order two. The transformation $\Delta (-1/\tau )={\tau }^{12}\Delta (\tau )$ givesThe Fricke involution exchanges infinity and zero, so $f$ has a zero of order two at the cusp zero.

Thus the morphism ${\varphi }_f:X_0(3)\to \hat{\mathbb{C}}$ has degree two, equal to its total pole order. An isomorphism of compact Riemann surfaces has degree one. Although $X_0(3)$ has genus zero, this particular morphism is therefore not an isomorphism.

If $p$ divides the primitive level $M$ of a trivial-character newform and $N=M{p}^r$, put $e_j=p^{jk/2}f(p^{j}z)$. Then $U_p{e}_0=a_p{e}_0$ and $U_p{e}_j=p^{k/2}{e}_{j-1}$ for $j\ge 1$. The normalized Fricke involution acts by $e_j\mapsto \epsilon e_{r-j}$, where $W_{M}f=\epsilon f$. These two operators act irreducibly: a nonzero invariant subspace contains an eigenvector of $U_p$; its eigenvalue is $a_p$ or zero. Either it contains $e_0$ immediately or applying the reversal and a power of $U_p$ produces $e_0$, using $a_p^2\ne p^k$. Reversal then produces $e_r$, and its successive images produce every basis vector.